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**Unformatted text preview: **Chapter 27
Current and Resistance
CHAPTE R OUTLI N E
27.1 Electric Current
27.2 Resistance
27.3 A Model for Electrical
Conduction
27.4 Resistance and Temperature
27.5 Superconductors
27.6 Electrical Power
L These power lines transfer energy from the power company to homes and businesses.
The energy is transferred at a very high voltage, possibly hundreds of thousands of volts in
some cases. Despite the fact that this makes power lines very dangerous, the high voltage
results in less loss of power due to resistance in the wires. (Telegraph Colour Library/ FPG)
831
Thus far our treatment of electrical phenomena has been conﬁned to the study of
charges in equilibrium situations, or electrostatics. We now consider situations involving
electric charges that are not in equilibrium. We use the term electric current, or simply
current, to describe the rate of ﬂow of charge through some region of space. Most practical applications of electricity deal with electric currents. For example, the battery in a
ﬂashlight produces a current in the ﬁlament of the bulb when the switch is turned on.
A variety of home appliances operate on alternating current. In these common situations, current exists in a conductor, such as a copper wire. It also is possible for
currents to exist outside a conductor. For instance, a beam of electrons in a television
picture tube constitutes a current.
This chapter begins with the deﬁnition of current. A microscopic description of current is given, and some of the factors that contribute to the opposition to the ﬂow of
charge in conductors are discussed. A classical model is used to describe electrical conduction in metals, and some of the limitations of this model are cited. We also deﬁne
electrical resistance and introduce a new circuit element, the resistor. We conclude by
discussing the rate at which energy is transferred to a device in an electric circuit.
27.1
+
+
+
+
A
+
I
Figure 27.1 Charges in motion
through an area A. The time rate at
which charge ﬂows through the
area is deﬁned as the current I.
The direction of the current is the
direction in which positive charges
ﬂow when free to do so.
Electric Current
In this section, we study the ﬂow of electric charges through a piece of material. The
amount of ﬂow depends on the material through which the charges are passing and the
potential difference across the material. Whenever there is a net ﬂow of charge through
some region, an electric current is said to exist.
It is instructive to draw an analogy between water ﬂow and current. In many
localities it is common practice to install low-ﬂow showerheads in homes as a waterconservation measure. We quantify the ﬂow of water from these and similar devices by
specifying the amount of water that emerges during a given time interval, which is
often measured in liters per minute. On a grander scale, we can characterize a river
current by describing the rate at which the water ﬂows past a particular location. For
example, the ﬂow over the brink at Niagara Falls is maintained at rates between
1 400 m3/s and 2 800 m3/s.
There is also an analogy between thermal conduction and current. In Section 20.7,
we discussed the ﬂow of energy by heat through a sample of material. The rate of
energy ﬂow is determined by the material as well as the temperature difference across
the material, as described by Equation 20.14.
To deﬁne current more precisely, suppose that charges are moving perpendicular to
a sur face of area A, as shown in Figure 27.1. (This area could be the cross-sectional area
of a wire, for example.) The current is the rate at which charge ﬂows through this
surface. If Q is the amount of charge that passes through this area in a time interval
t, the average current I av is equal to the charge that passes through A per unit time:
Iav
832
∆Q
∆t
(27.1)
S E C T I O N 2 7. 1 • Electric Current
833
If the rate at which charge ﬂows varies in time, then the current varies in time; we
deﬁne the instantaneous current I as the differential limit of average current:
dQ
dt
I
(27.2)
Electric current
The SI unit of current is the ampere (A):
1A
1C
1s
(27.3)
That is, 1 A of current is equivalent to 1 C of charge passing through the sur face area
in 1 s.
The charges passing through the sur face in Figure 27.1 can be positive or negative,
or both. It is conventional to assign to the current the same direction as the ﬂow
of positive charge. In electrical conductors, such as copper or aluminum, the current
is due to the motion of negatively charged electrons. Therefore, when we speak of
current in an ordinary conductor, the direction of the current is opposite the
direction of ﬂow of electrons. However, if we are considering a beam of positively
charged protons in an accelerator, the current is in the direction of motion of the
protons. In some cases—such as those involving gases and electrolytes, for instance—
the current is the result of the ﬂow of both positive and negative charges.
If the ends of a conducting wire are connected to form a loop, all points on the
loop are at the same electric potential, and hence the electric ﬁeld is zero within and
at the sur face of the conductor. Because the electric ﬁeld is zero, there is no net
transport of charge through the wire, and therefore there is no current. However, if
the ends of the conducting wire are connected to a battery, all points on the loop are
not at the same potential. The battery sets up a potential difference between the ends
of the loop, creating an electric ﬁeld within the wire. The electric ﬁeld exerts forces on
the conduction electrons in the wire, causing them to move in the wire, thus creating a
current.
It is common to refer to a moving charge (positive or negative) as a mobile charge
carrier. For example, the mobile charge carriers in a metal are electrons.
We can relate current to the motion of the charge carriers by describing a microscopic
model of conduction in a metal. Consider the current in a conductor of cross-sectional
area A (Fig. 27.2). The volume of a section of the conductor of length x (the gray
region shown in Fig. 27.2) is A x. If n represents the number of mobile charge
carriers per unit volume (in other words, the charge carrier density), the number of
carriers in the gray section is nA x. Therefore, the total charge Q in this section is
number of carriers in section
charge per carrier
(nA x)q
where q is the charge on each carrier. If the carriers move with a speed vd , the displacement they experience in the x direction in a time interval t is x vd t. Let us
choose t to be the time interval required for the charges in the cylinder to move
through a displacement whose magnitude is equal to the length of the cylinder. This
time interval is also that required for all of the charges in the cylinder to pass through
the circular area at one end. With this choice, we can write Q in the form
Q
27.1 “Current Flow” Is
Redundant
The phrase current ﬂow is commonly used, although it is strictly
incorrect, because current is a
ﬂow (of charge). This is similar
to the phrase heat transfer, which
is also redundant because heat is
a transfer (of energy). We will
avoid this phrase and speak of
ﬂow of charge or charge ﬂow .
∆x
vd
A
vd ∆t
Figure 27.2 A section of a uniform
conductor of cross-sectional area A.
The mobile charge carriers move
with a speed vd , and the displacement they experience in the x
direction in a time interval t is
x vd t. If we choose t to
be the time interval during which
the charges are displaced, on
the average, by the length of the
cylinder, the number of carriers in
the section of length x is n Avd t ,
where n is the number of carriers
per unit volume.
(nAvd t)q
If we divide both sides of this equation by
conductor is
I av
PITFALL PR EVE NTI O N
q
Microscopic Model of Current
Q
L
∆Q
∆t
t, we see that the average current in the
nqvdA
(27.4)
Current in a conductor in terms
of microscopic quantities
834
C HAPTE R 2 7 • Current and Resistance
vd
–
E
Figure 27.3 A schematic
representation of the zigzag
motion of an electron in a
conductor. The changes in
direction are the result of collisions
between the electron and atoms in
the conductor. Note that the net
motion of the electron is opposite
the direction of the electric ﬁeld.
Because of the acceleration of the
charge carriers due to the electric
force, the paths are actually
parabolic. However, the drift speed
is much smaller than the average
speed, so the parabolic shape is not
visible on this scale.
The speed of the charge carriers vd is an average speed called the drift speed. To
understand the meaning of drift speed, consider a conductor in which the charge carriers are free electrons. If the conductor is isolated—that is, the potential difference
across it is zero—then these electrons undergo random motion that is analogous to the
motion of gas molecules. As we discussed earlier, when a potential difference is applied
across the conductor (for example, by means of a battery), an electric ﬁeld is set up in
the conductor; this ﬁeld exerts an electric force on the electrons, producing a current.
However, the electrons do not move in straight lines along the conductor. Instead, they
collide repeatedly with the metal atoms, and their resultant motion is complicated and
zigzag (Fig. 27.3). Despite the collisions, the electrons move slowly along the conductor (in a direction opposite that of E) at the drift velocity vd .
We can think of the atom–electron collisions in a conductor as an effective internal
friction (or drag force) similar to that experienced by the molecules of a liquid ﬂowing
through a pipe stuffed with steel wool. The energy transferred from the electrons to
the metal atoms during collisions causes an increase in the vibrational energy of the
atoms and a corresponding increase in the temperature of the conductor.
Quick Quiz 27.1
Consider positive and negative charges moving horizontally through the four regions shown in Figure 27.4. Rank the current in these four
regions, from lowest to highest.
+
+
–
+
+
–
+
–
+
+
+
+
(a)
(b)
–
–
–
(c)
(d)
Figure 27.4 (Quick Quiz 27.1) Charges move through four regions.
Quick Quiz 27.2
Electric charge is conserved. As a consequence, when current
arrives at a junction of wires, the charges can take either of two paths out of the junction
and the numerical sum of the currents in the two paths equals the current that entered
the junction. Thus, current is (a) a vector (b) a scalar (c) neither a vector nor a scalar.
Example 27.1
Drift Speed in a Copper Wire
The 12-gauge copper wire in a typical residential building
has a cross-sectional area of 3.31 10 6 m2. If it carries a
current of 10.0 A, what is the drift speed of the electrons?
Assume that each copper atom contributes one free
electron to the current. The density of copper is 8.95 g/cm3.
Solution From the periodic table of the elements in
Appendix C, we ﬁnd that the molar mass of copper is
63.5 g/mol. Recall that 1 mol of any substance contains
Avogadro’s number of atoms (6.02 1023). Knowing the
density of copper, we can calculate the volume occupied by
63.5 g ( 1 mol) of copper:
V
m
63.5 g
8.95 g/cm3
7.09 cm3
Because each copper atom contributes one free electron to
the current, we have
6.02
1023 electrons
7.09 cm3
8.49
n
1028 electrons/m3
1.00
106 cm3
1 m3
From Equation 27.4, we ﬁnd that the drift speed is
I
nqA
vd
where q is the absolute value of the charge on each electron.
Thus,
vd
I
nqA
(8.49
2.22
1028 m
10
4
10.0 C/s
10 19 C)(3.31
3)(1.60
m/s
10
6
m2)
S E C T I O N 2 7. 2 • Resistance
Example 27.1 shows that typical drift speeds are ver y low. For instance, electrons
traveling with a speed of 2.22 10 4 m/s would take about 75 min to travel 1 m! In
view of this, you might wonder why a light turns on almost instantaneously when a
switch is thrown. In a conductor, changes in the electric ﬁeld that drives the free
electrons travel through the conductor with a speed close to that of light. Thus,
when you ﬂip on a light switch, electrons already in the ﬁlament of the lightbulb
experience electric forces and begin moving after a time inter val on the order of
nanoseconds.
27.2 Resistance
In Chapter 24 we found that the electric ﬁeld inside a conductor is zero. However, this
statement is true only if the conductor is in static equilibrium. The purpose of this
section is to describe what happens when the charges in the conductor are not in equilibrium, in which case there is an electric ﬁeld in the conductor.
Consider a conductor of cross-sectional area A carrying a current I. The current
density J in the conductor is deﬁned as the current per unit area. Because the current
I nqvd A, the current density is
I
A
J
nqvd
L
835
PITFALL PR EVE NTI O N
27.2 Electrons Are
Available Everywhere
Electrons do not have to travel from
the light switch to the light in order
for the light to operate. Electrons
already in the ﬁlament of the
lightbulb move in response to the
electric ﬁeld set up by the battery.
Notice also that a battery does
not provide electrons to the
circuit. It establishes the electric
ﬁeld that exerts a force on
electrons already in the wires and
elements of the circuit.
(27.5)
where J has SI units of A/m2. This expression is valid only if the current density is uniform and only if the sur face of cross-sectional area A is perpendicular to the direction
of the current. In general, current density is a vector quantity:
J
nq vd
(27.6)
Current density
From this equation, we see that current density is in the direction of charge motion
for positive charge carriers and opposite the direction of motion for negative
charge carriers.
A current density J and an electric ﬁeld E are established in a conductor
whenever a potential difference is maintained across the conductor. In some
materials, the current density is proportional to the electric ﬁeld:
J
E
(27.7)
where the constant of proportionality is called the conductivity of the conductor.1
Materials that obey Equation 27.7 are said to follow Ohm’s law, named after Georg
Simon Ohm (1789–1854). More speciﬁcally, Ohm’s law states that
for many materials (including most metals), the ratio of the current density to the
electric ﬁeld is a constant that is independent of the electric ﬁeld producing the
current.
Materials that obey Ohm’s law and hence demonstrate this simple relationship
between E and J are said to be ohmic. Experimentally, however, it is found that not all
materials have this property. Materials and devices that do not obey Ohm’s law are said
to be nonohmic. Ohm’s law is not a fundamental law of nature but rather an empirical
relationship valid only for certain materials.
We can obtain an equation useful in practical applications by considering a
segment of straight wire of uniform cross-sectional area A and length , as shown in
1
Do not confuse conductivity
with sur face charge density, for which the same symbol is used.
Georg Simon Ohm
German physicist (1789–1854)
Ohm, a high school teacher and
later a professor at the University
of Munich, formulated the
concept of resistance and
discovered the proportionalities
expressed in Equations 27.7 and
27.8. (© Bettmann/Corbis)
836
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C HAPTE R 2 7 • Current and Resistance
PITFALL PR EVE NTI O N
27.3 We’ve Seen
Something Like
Equation 27.8 Before
In Chapter 5, we introduced
Newton’s second law, F ma,
for a net force on an object of
mass m. This can be written as
m
F
a
In that chapter, we deﬁned mass
as resistance to a change in motion in
response to an external force. Mass as
resistance to changes in motion is
analogous to electrical resistance
to charge ﬂow, and Equation 27.8
is analogous to the form of
Newton’s second law shown here.
A
Vb
Va
E
Figure 27.5 A uniform conductor of length and cross-sectional area A. A potential
difference V Vb Va maintained across the conductor sets up an electric ﬁeld E,
and this ﬁeld produces a current I that is proportional to the potential difference.
Figure 27.5. A potential difference V Vb Va is maintained across the wire, creating in the wire an electric ﬁeld and a current. If the ﬁeld is assumed to be uniform, the
potential difference is related to the ﬁeld through the relationship2
V
J
Many individuals call Equation
27.8 Ohm’s law, but this is incorrect. This equation is simply the
deﬁnition of resistance, and provides an important relationship
between voltage, current, and
resistance. Ohm’s law is related
to a linear relationship between J
and E (Eq. 27.7) or, equivalently,
between I and V, which, from
Equation 27.8, indicates that the
resistance is constant, independent of the applied voltage.
∆V
E
I/A, we can write the potential difference as
∆V
PITFALL PR EVE NTI O N
27.4 Equation 27.8 Is Not
Ohm’s Law
E
Therefore, we can express the magnitude of the current density in the wire as
Because J
L
I
J
I
A
RI
The quantity R
/ A is called the resistance of the conductor. We can deﬁne the
resistance as the ratio of the potential difference across a conductor to the current in
the conductor:
∆V
I
R
(27.8)
We will use this equation over and over again when studying electric circuits. From this
result we see that resistance has SI units of volts per ampere. One volt per ampere is
deﬁned to be one ohm ( ):
1V
1A
1
(27.9)
This expression shows that if a potential difference of 1 V across a conductor causes a
current of 1 A, the resistance of the conductor is 1 . For example, if an electrical
appliance connected to a 120-V source of potential difference carries a current of 6 A,
its resistance is 20 .
The inverse of conductivity is resistivity3 :
1
Resistivity is the inverse of
conductivity
(27.10)
where has the units ohm-meters ( m). Because R
resistance of a uniform block of material along the length
Resistance of a uniform
material along the length
R
2
(27.11)
A
This result follows from the deﬁnition of potential difference:
Vb
3
/ A, we can express the
as
Do not confuse resistivity
Va
b
a
E ds
E
0
dx
E
with mass density or charge density, for which the same symbol is used.
S E C T I O N 2 7. 2 • Resistance
L
Table 27.1
Resistivities and Temperature Coefﬁcients of Resistivity
for Various Materials
Material
Resistivitya(
Silver
Copper
Gold
Aluminum
Tungsten
Iron
Platinum
Lead
Nichromec
Carbon
Germanium
Silicon
Glass
Hard rubber
Sulfur
Quartz (fused)
1.59 10 8
1.7 10 8
2.44 10 8
2.82 10 8
5.6 10 8
10 10 8
11 10 8
22 10 8
1.50 10 6
3.5 10 5
0.46
640
1010 to 1014
1013
1015
75 1016
m)
Temperature
Coefﬁcientb [( C)
3.8
3.9
3.4
3.9
4.5
5.0
3.92
3.9
0.4
0.5
48
75
10
10
10
10
10
10
10
10
10
10
10
10
1]
3
3
3
3
3
3
3
3
3
3
3
All values at 20°C.
b
See Section 27.4.
c
A nickel–chromium alloy commonly used in heating elements.
Henry Leap and Jim Lehman
Every ohmic material has a characteristic resistivity that depends on the properties of
the material and on temperature. Additionally, as you can see from Equation 27.11, the
resistance of a sample depends on geometry as well as on resistivity. Table 27.1 gives
the resistivities of a variety of materials at 20°C. Note the enormous range, from very
low values for good conductors such as copper and silver, to very high values for good
insulators such as glass and rubber. An ideal conductor would have zero resistivity, and
an ideal insulator would have inﬁnite resistivity.
Equation 27.11 shows that the resistance of a given cylindrical conductor such as a
wire is proportional to its length and inversely proportional to its cross-sectional area.
If the length of a wire is doubled, then its resistance doubles. If its cross-sectional area
is doubled, then its resistance decreases by one half. The situation is analogous to the
ﬂow of a liquid through a pipe. As the pipe’s length is increased, the resistance to ﬂow
increases. As the pipe’s cross-sectional area is increased, more liquid crosses a given
cross section of the pipe per unit time interval. Thus, more liquid ﬂows for the same
pressure differential applied to the pipe, and the resistance to ﬂow decreases.
An assortment of resistors used in electrical circuits.
PITFALL PR EVE NTI O N
27.5 Resistance and
Resistivity
3
a
837
Resistivity is property of a substance, while resistance is a property of an object. We have seen
similar pairs of variables before.
For example, density is a property of a substance, while mass is
a property of an object. Equation
27.11 relates resistance to resistivity, and we have seen a previous
equation (Equation 1.1) which
relates mass to density.
838
C HAPTE R 2 7 • Current and Resistance
Table 27.2
Color Coding for Resistors
Figure 27.6 The colored bands on a resistor represent a code for
determining resistance. The ﬁrst two colors give the ﬁrst two digits in the
resistance value. The third color represents the power of ten for the
multiplier of the resistance value. The last color is the tolerance of the
resistance value. As an example, the four colors on the circled resistors are
red ( 2), black ( 0), orange ( 103), and gold ( 5%), and so the
resistance value is 20 103
20 k with a tolerance value of 5% 1 k .
(The values for the colors are from Table 27.2.)
Number
Multiplier
Black
Brown
Red
Orange
Yellow
Green
Blue
Violet
Gray
White
Gold
Silver
Colorless
SuperStock
Color
0
1
2
3
4
5
6
7
8
9
1
101
102
103
104
105
106
107
108
109
10
10
1
2
Tolerance
5%
10%
20%
Most electric circuits use circuit elements called resistors to control the current
level in the various parts of the circuit. Two common types of resistors are the composition resistor, which contains carbon, and the wire-wound resistor, which consists of a coil of
wire. Values of resistors in ohms are normally indicated by color-coding, as shown in
Figure 27.6 and Table 27.2.
Ohmic materials and devices have a linear current–potential difference relationship over a broad range of applied potential differences (Fig. 27.7a). The slope of the
I-versus- V curve in the linear region yields a value for 1/R. Nonohmic materials have
a nonlinear current–potential difference relationship. One common semiconducting
device that has nonlinear I-versus- V characteristics is the junction diode (Fig. 27.7b).
The resistance of this device is low for currents in one direction (positive V ) and
high for currents in the reverse direction (negative V ). In fact, most modern
electronic devices, such as transistors, have nonlinear current–potential difference
relationships; their proper operation depends on the particular way in which they
violate Ohm’s law.
Quick Quiz 27.3
Suppose that a current-carrying ohmic metal wire has a
cross-sectional area that gradually becomes smaller from one end of the wire to the
other. The current must have the same value in each section of the wire so that charge
does not accumulate at any one point. How do the drift velocity and the resistance per
I
I
Slope = 1
R
V
(a)
V
(b)
Figure 27.7 (a) The current–potential difference curve for an ohmic material. The
curve is linear, and the slope is equal to the inverse of the resistance of the conductor.
(b) A nonlinear current–potential difference curve for a junction diode. This device
does not obey Ohm’s law.
S E C T I O N 2 7. 2 • Resistance
839
unit length vary along the wire as the area becomes smaller? (a) The drift velocity and
resistance both increase. (b) The drift velocity and resistance both decrease. (c) The
drift velocity increases and the resistance decreases. (d) The drift velocity decreases
and the resistance increases.
Quick Quiz 27.4
A cylindrical wire has a radius r and length . If both r and
are doubled, the resistance of the wire (a) increases (b) decreases (c) remains the same.
Quick Quiz 27.5 In Figure 27.7b, as the applied voltage increases, the resistance of the diode (a) increases (b) decreases (c) remains the same.
Example 27.2
The Resistance of a Conductor
Calculate the resistance of an aluminum cylinder that
has a length of 10.0 cm and a cross-sectional area of
2.00 10 4 m2. Repeat the calculation for a cylinder of the
same dimensions and made of glass having a resistivity of
3.0 1010
m.
Solution From Equation 27.11 and Table 27.1, we can calculate the resistance of the aluminum cylinder as follows:
R
(2.82
A
1.41
10
Example 27.3
10
8
m)
0.100 m
2.00 10 4 m2
5
Solution The cross-sectional area of this wire is
A
(0.321
A
(3.0
A
1010
m)
0.100 m
2.00 10 4 m2
1013
1.5
As you might guess from the large difference in resistivities,
the resistances of identically shaped cylinders of aluminum
and glass differ widely. The resistance of the glass cylinder is
18 orders of magnitude greater than that of the aluminum
cylinder.
Interactive
10
3
m)2
3.24
10
1.5 10 6
m
7 m2
3.24 10
4.6
Solution Because a 1.0-m length of this wire has a resistance of 4.6 , Equation 27.8 gives
I
7
m2
The resistivity of Nichrome is 1.5 10 6
m (see
Table 27.1). Thus, we can use Equation 27.11 to ﬁnd the
resistance per unit length:
R
R
The Resistance of Nichrome Wire
(A) Calculate the resistance per unit length of a 22-gauge
Nichrome wire, which has a radius of 0.321 mm.
r2
Similarly, for glass we ﬁnd that
/m
(B) If a potential difference of 10 V is maintained across a
1.0-m length of the Nichrome wire, what is the current in
the wire?
∆V
R
10 V
4.6
2.2 A
Note from Table 27.1 that the resistivity of Nichrome wire is
about 100 times that of copper. A copper wire of the same
radius would have a resistance per unit length of only
0.052 /m. A 1.0-m length of copper wire of the same
radius would carry the same current (2.2 A) with an applied
potential difference of only 0.11 V.
Because of its high resistivity and its resistance to
oxidation, Nichrome is often used for heating elements in
toasters, irons, and electric heaters.
Explore the resistance of different materials at the Interactive Worked Example link at http://www.pse6.com.
Example 27.4
The Radial Resistance of a Coaxial Cable
Coaxial cables are used extensively for cable television and
other electronic applications. A coaxial cable consists of
two concentric cylindrical conductors. The region
between the conductors is completely ﬁlled with silicon, as
shown in Figure 27.8a, and current leakage through the
silicon, in the radial direction, is unwanted. (The cable is
designed to conduct current along its length—this is not
the current we are considering here.) The radius of the
inner conductor is a 0.500 cm, the radius of the outer
one is b 1.75 cm, and the length is L 15.0 cm.
840
C HAPTE R 2 7 • Current and Resistance
L
Cur rent
direction
dr
Silicon
a
r
b
Inner
conductor
Outer
conductor
(a)
End view
(b)
Figure 27.8 (Example 27.4) A coaxial cable. (a) Silicon ﬁlls the gap between the two
conductors. (b) End view, showing current leakage.
Calculate the resistance of the silicon between the two
conductors.
Solution Conceptualize by imagining two currents, as
suggested in the text of the problem. The desired current is
along the cable, carried within the conductors. The
undesired current corresponds to charge leakage through
the silicon and its direction is radial. Because we know the
resistivity and the geometry of the silicon, we categorize this
as a problem in which we ﬁnd the resistance of the silicon
from these parameters, using Equation 27.11. Because the
area through which the charges pass depends on the radial
position, we must use integral calculus to determine the
answer.
To analyze the problem, we divide the silicon into concentric elements of inﬁnitesimal thickness dr (Fig. 27.8b).
We start by using the differential form of Equation 27.11,
replacing
with r for the distance variable: dR
dr/A,
where dR is the resistance of an element of silicon of
thickness dr and sur face area A. In this example, we take as
our representative concentric element a hollow silicon cylinder of radius r, thickness dr, and length L, as in Figure 27.8.
Any charge that passes from the inner conductor to the
outer one must pass radially through this concentric element, and the area through which this charge passes is
A 2 rL. (This is the curved sur face area—circumference
multiplied by length—of our hollow silicon cylinder of
thickness dr.) Hence, we can write the resistance of our
hollow cylinder of silicon as
dR
dr
2 rL
Because we wish to know the total resistance across the
entire thickness of the silicon, we must integrate this expression from r a to r b :
(1)
R
b
a
dR
b
2L
a
dr
r
2L
ln
b
a
Substituting in the values given, and using
silicon, we obtain
640
m
ln
2 (0.150 m)
R
1.75 cm
0.500 cm
640
m for
851
To ﬁnalize this problem, let us compare this resistance to
that of the inner conductor of the cable along the 15.0-cm
length. Assuming that the conductor is made of copper, we
have
R
A
3.2
(1.7
10
10
8
m)
0.150 m
(5.00 10 3 m)2
5
This resistance is much smaller than the radial resistance. As
a consequence, almost all of the current corresponds to
charge moving along the length of the cable, with a very
small fraction leaking in the radial direction.
What If? Suppose the coaxial cable is enlarged to twice the
overall diameter with two possibilities: (1) the ratio b/a is
held ﬁxed, or (2) the difference b a is held ﬁxed. For which
possibility does the leakage current between the inner and
outer conductors increase when the voltage is applied
between the two conductors?
Answer In order for the current to increase, the resistance
must decrease. For possibility (1), in which b/a is held ﬁxed,
Equation (1) tells us that the resistance is unaffected. For
possibility (2), we do not have an equation involving the difference b a to inspect. Looking at Figure 27.8b, however,
we see that increasing b and a while holding the voltage
constant results in charge ﬂowing through the same
thickness of silicon but through a larger overall area perpendicular to the ﬂow. This larger area will result in lower resistance and a higher current.
S E C T I O N 2 7. 3 • A Model for Electrical Conduction
27.3
841
A Model for Electrical Conduction
In this section we describe a classical model of electrical conduction in metals that was
ﬁrst proposed by Paul Drude (1863–1906) in 1900. This model leads to Ohm’s law and
shows that resistivity can be related to the motion of electrons in metals. Although the
Drude model described here does have limitations, it nevertheless introduces concepts
that are still applied in more elaborate treatments.
Consider a conductor as a regular array of atoms plus a collection of free electrons,
which are sometimes called conduction electrons. The conduction electrons, although
bound to their respective atoms when the atoms are not part of a solid, gain mobility
when the free atoms condense into a solid. In the absence of an electric ﬁeld, the conduction electrons move in random directions through the conductor with average speeds on
the order of 106 m/s. The situation is similar to the motion of gas molecules conﬁned in a
vessel. In fact, some scientists refer to conduction electrons in a metal as an electron gas.
There is no current in the conductor in the absence of an electric ﬁeld because the drift
velocity of the free electrons is zero. That is, on the average, just as many electrons move
in one direction as in the opposite direction, and so there is no net ﬂow of charge.
This situation changes when an electric ﬁeld is applied. Now, in addition to undergoing the random motion just described, the free electrons drift slowly in a direction
opposite that of the electric ﬁeld, with an average drift speed vd that is much smaller
(typically 10 4 m/s) than their average speed between collisions (typically 106 m/s).
Figure 27.9 provides a crude description of the motion of free electrons in a
conductor. In the absence of an electric ﬁeld, there is no net displacement after many
collisions (Fig. 27.9a). An electric ﬁeld E modiﬁes the random motion and causes the
electrons to drift in a direction opposite that of E (Fig. 27.9b).
In our model, we assume that the motion of an electron after a collision is independent of its motion before the collision. We also assume that the excess energy acquired
by the electrons in the electric ﬁeld is lost to the atoms of the conductor when the
electrons and atoms collide. The energy given up to the atoms increases their vibrational energy, and this causes the temperature of the conductor to increase. The temperature increase of a conductor due to resistance is utilized in electric toasters and
other familiar appliances.
We are now in a position to derive an expression for the drift velocity. When a free
e) is subjected to an electric ﬁeld E, it
electron of mass me and charge q (
experiences a force F q E. Because this force is related to the acceleration of the
electron through Newton’s second law, F me a, we conclude that the acceleration of
the electron is
qE
me
a
(27.12)
E
––
–
–
–
–
–
–
(a)
(b)
Active Figure 27.9 (a) A schematic diagram of the random motion of two charge
carriers in a conductor in the absence of an electric ﬁeld. The drift velocity is zero.
(b) The motion of the charge carriers in a conductor in the presence of an electric
ﬁeld. Note that the random motion is modiﬁed by the ﬁeld, and the charge carriers
have a drift velocity.
At the Active Figures link
at http://www.pse6.com, you
can adjust the electric ﬁeld to
see the resulting effect on the
motion of an electron.
842
C HAPTE R 2 7 • Current and Resistance
This acceleration, which occurs for only a short time interval between collisions,
enables the electron to acquire a small drift velocity. If vi is the electron’s initial velocity
the instant after a collision (which occurs at a time that we deﬁne as t 0), then the
velocity of the electron at time t (at which the next collision occurs) is
vf
vi
at
qE
t
me
vi
(27.13)
We now take the average value of vf over all possible collision times t and all possible
values of vi . If we assume that the initial velocities are randomly distributed over all
possible values, we see that the average value of vi is zero. The term (q E/me)t is the
velocity change of the electron due to the electric ﬁeld during one trip between atoms.
The average value of the second term of Equation 27.13 is (q E/me ) , where is the
average time inter val between successive collisions. Because the average value of vf is equal to
the drift velocity, we have
Drift velocity in terms of
microscopic quantities
vf
qE
me
vd
(27.14)
We can relate this expression for drift velocity to the current in the conductor.
Substituting Equation 27.14 into Equation 27.6, we ﬁnd that the magnitude of the
current density is
J
Current density in terms of
microscopic quantities
nq 2E
me
nqvd
(27.15)
where n is the number of charge carriers per unit volume. Comparing this expression
with Ohm’s law, J
E , we obtain the following relationships for conductivity and
resistivity of a conductor:
nq 2
me
Conductivity in terms of microscopic quantities
(27.16)
1
Resistivity in terms of microscopic quantities
me
nq 2
(27.17)
According to this classical model, conductivity and resistivity do not depend on the
strength of the electric ﬁeld. This feature is characteristic of a conductor obeying
Ohm’s law.
The average time interval between collisions is related to the average distance
between collisions (that is, the mean free path; see Section 21.7) and the average speed
v through the expression
(27.18)
v
Example 27.5
Electron Collisions in a Wire
(A) Using the data and results from Example 27.1 and the
classical model of electron conduction, estimate the average
time interval between collisions for electrons in household
copper wiring.
Solution From Equation 27.17, we see that
me
nq 2
where
sity is n
1.7 10 8
m for copper and the carrier den8.49 1028 electrons/m3 for the wire described
in Example 27.1. Substitution of these values into the
expression above gives
9.11
(8.49
2.5
1028
10
14
m
3)(1.6
10 31 kg
10 19 C)2 (1.7
10
8
m)
s
(B) Assuming that the average speed for free electrons in
copper is 1.6 106 m/s and using the result from part (A),
calculate the mean free path for electrons in copper.
S E C T I O N 2 7. 4 • Resistance and Temperature
Solution From Equation 27.18,
v
(1.6
4.0
10
106 m/s)(2.5
8
10
14
843
which is equivalent to 40 nm (compared with atomic
spacings of about 0.2 nm). Thus, although the time interval
between collisions is very short, an electron in the wire
travels about 200 atomic spacings between collisions.
s)
m
27.4 Resistance and Temperature
Over a limited temperature range, the resistivity of a conductor varies approximately
linearly with temperature according to the expression
0[1
(T
T0)]
(27.19)
Variation of
with temperature
where is the resistivity at some temperature T (in degrees Celsius), 0 is the resistivity
at some reference temperature T0 (usually taken to be 20°C), and is the temperature coefﬁcient of resistivity. From Equation 27.19, we see that the temperature
coefﬁcient of resistivity can be expressed as
1∆
0 ∆T
(27.20)
Temperature coefﬁcient of
resistivity
where
T T0.
0 is the change in resistivity in the temperature interval T
The temperature coefﬁcients of resistivity for various materials are given in
Table 27.1. Note that the unit for is degrees Celsius 1 [(°C) 1]. Because resistance
is proportional to resistivity (Eq. 27.11), we can write the variation of resistance as
R
R 0[1
(T
T0)]
(27.21)
Use of this property enables us to make precise temperature measurements, as shown
in Example 27.6.
Quick Quiz 27.6 When does a lightbulb carry more current: (a) just after it
is turned on and the glow of the metal ﬁlament is increasing, or (b) after it has been
on for a few milliseconds and the glow is steady?
Example 27.6
A Platinum Resistance Thermometer
A resistance thermometer, which measures temperature by
measuring the change in resistance of a conductor, is made
from platinum and has a resistance of 50.0
at 20.0°C.
When immersed in a vessel containing melting indium, its
resistance increases to 76.8 . Calculate the melting point
of the indium.
∆T
R
R0
R0
[3.92
76.8
10
3(
50.0
C ) 1](50.0
)
137 C
Because T0
20.0°C, we ﬁnd that T, the temperature of the
melting indium sample, is 157 C.
Solution Solving Equation 27.21 for T and using the
value for platinum given in Table 27.1, we obtain
For metals like copper, resistivity is nearly proportional to temperature, as shown
in Figure 27.10. However, a nonlinear region always exists at very low temperatures,
and the resistivity usually reaches some ﬁnite value as the temperature approaches
absolute zero. This residual resistivity near absolute zero is caused primarily by the
C HAPTE R 2 7 • Current and Resistance
844
ρ
0
T
ρ
ρ0
0
collision of electrons with impurities and imper fections in the metal. In contrast, hightemperature resistivity (the linear region) is predominantly characterized by collisions
between electrons and metal atoms.
Notice that three of the values in Table 27.1 are negative; this indicates that the
resistivity of these materials decreases with increasing temperature (Fig. 27.11), which
is indicative of a class of materials called semiconductors. This behavior is due to an
increase in the density of charge carriers at higher temperatures.
Because the charge carriers in a semiconductor are often associated with impurity
atoms, the resistivity of these materials is very sensitive to the type and concentration of
such impurities. We shall return to the study of semiconductors in Chapter 43.
27.5
T
Figure 27.10 Resistivity versus
temperature for a metal such as
copper. The curve is linear over a
wide range of temperatures, and
increases with increasing
temperature. As T approaches
absolute zero (inset), the resistivity
approaches a ﬁnite value 0.
ρ
Superconductors
There is a class of metals and compounds whose resistance decreases to zero when they
are below a certain temperature Tc , known as the critical temperature. These
materials are known as superconductors. The resistance–temperature graph for a
superconductor follows that of a normal metal at temperatures above Tc (Fig. 27.12).
When the temperature is at or below Tc , the resistivity drops suddenly to zero. This
phenomenon was discovered in 1911 by the Dutch physicist Heike Kamerlingh-Onnes
(1853–1926) as he worked with mercury, which is a superconductor below 4.2 K.
Recent measurements have shown that the resistivities of superconductors below their
m—around 1017 times smaller than the resistivity
Tc values are less than 4 10 25
of copper and in practice considered to be zero.
Today thousands of superconductors are known, and as Table 27.3 illustrates, the
critical temperatures of recently discovered superconductors are substantially higher
than initially thought possible. Two kinds of superconductors are recognized. The
more recently identiﬁed ones are essentially ceramics with high critical temperatures,
whereas superconducting materials such as those observed by Kamerlingh-Onnes are
metals. If a room-temperature superconductor is ever identiﬁed, its impact on
technology could be tremendous.
The value of Tc is sensitive to chemical composition, pressure, and molecular
structure. It is interesting to note that copper, silver, and gold, which are excellent
conductors, do not exhibit superconductivity.
T
Figure 27.11 Resistivity versus
temperature for a pure
semiconductor, such as silicon or
germanium.
R(Ω)
0.15
Table 27.3
Critical Temperatures for
Various Superconductors
0.10
Material
0.05
Tc
0.00
4.0
4.2
T (K)
4.4
Figure 27.12 Resistance versus temperature for a
sample of mercury (Hg). The graph follows that of a
normal metal above the critical temperature Tc . The
resistance drops to zero at Tc , which is 4.2 K for mercury.
Tc(K)
HgBa2Ca2Cu3O8
Tl–Ba–Ca–Cu–O
Bi–Sr–Ca–Cu–O
YBa2Cu3O7
Nb3Ge
Nb3Sn
Nb
Pb
Hg
Sn
Al
Zn
134
125
105
92
23.2
18.05
9.46
7.18
4.15
3.72
1.19
0.88
One of the truly remarkable features of superconductors is that once a current is
set up in them, it persists without any applied potential difference (because R 0). Steady
currents have been observed to persist in superconducting loops for several years with
no apparent decay!
An important and useful application of superconductivity is in the development of
superconducting magnets, in which the magnitudes of the magnetic ﬁeld are about
ten times greater than those produced by the best normal electromagnets. Such
superconducting magnets are being considered as a means of storing energy.
Superconducting magnets are currently used in medical magnetic resonance imaging
(MRI) units, which produce high-quality images of internal organs without the need
for excessive exposure of patients to x-rays or other harmful radiation.
For further information on superconductivity, see Section 43.8.
27.6
Electrical Power
If a battery is used to establish an electric current in a conductor, there is a continuous
transformation of chemical energy in the battery to kinetic energy of the electrons to
internal energy in the conductor, resulting in an increase in the temperature of the
conductor.
In typical electric circuits, energy is transferred from a source such as a battery, to
some device, such as a lightbulb or a radio receiver. Let us determine an expression
that will allow us to calculate the rate of this energy transfer. First, consider the simple
circuit in Figure 27.13, where we imagine energy is being delivered to a resistor.
(Resistors are designated by the circuit symbol
.) Because the connecting
wires also have resistance, some energy is delivered to the wires and some energy to the
resistor. Unless noted otherwise, we shall assume that the resistance of the wires is so
small compared to the resistance of the circuit element that we ignore the energy
delivered to the wires.
Imagine following a positive quantity of charge Q that is moving clockwise around
the circuit in Figure 27.13 from point a through the battery and resistor back to point a.
We identify the entire circuit as our system. As the charge moves from a to b through
the battery, the electric potential energy of the system increases by an amount Q V while
the chemical potential energy in the battery decreases by the same amount. (Recall from
Eq. 25.9 that U q V.) However, as the charge moves from c to d through the
resistor, the system loses this electric potential energy during collisions of electrons with
atoms in the resistor. In this process, the energy is transformed to internal energy
corresponding to increased vibrational motion of the atoms in the resistor. Because we
have neglected the resistance of the interconnecting wires, no energy transformation
occurs for paths bc and da. When the charge returns to point a, the net result is that
some of the chemical energy in the battery has been delivered to the resistor and
resides in the resistor as internal energy associated with molecular vibration.
The resistor is normally in contact with air, so its increased temperature will result
in a transfer of energy by heat into the air. In addition, the resistor emits thermal
845
Courtesy of IBM Research Laboratory
S E C T I O N 2 7. 6 • Electrical Power
A small permanent magnet
levitated above a disk of the
superconductor YBa2Cu3O7, which
is at 77 K.
L
PITFALL PR EVE NTI O N
27.6 Misconceptions
About Current
There are several common
misconceptions associated with
current in a circuit like that in
Figure 27.13. One is that current
comes out of one terminal of the
battery and is then “used up” as it
passes through the resistor,
leaving current in only one part
of the circuit. The truth is that
the current is the same ever ywhere
in the circuit. A related misconception has the current coming
out of the resistor being smaller
than that going in, because some
of the current is “used up.”
Another
misconception
has
current coming out of both
terminals of the battery, in
opposite directions, and then
“clashing” in the resistor, delivering the energy in this manner.
This is not the case—the charges
ﬂow in the same rotational sense
at all points in the circuit.
I
b
+
–
a
c
∆V
R
d
Active Figure 27.13 A circuit consisting of a resistor of
resistance R and a battery having a potential difference V across
its terminals. Positive charge ﬂows in the clockwise direction.
At the Active Figures link
at http://www.pse6.com, you
can adjust the battery voltage
and the resistance to see the
resulting current in the circuit
and power delivered to the
resistor.
846
L
C HAPTE R 2 7 • Current and Resistance
PITFALL PR EVE NTI O N
27.7 Charges Do Not
Move All the Way
Around a Circuit in a
Short Time
Due to the very small magnitude
of the drift velocity, it might take
hours for a single electron to
make one complete trip around
the circuit. In terms of understanding the energy transfer in a
circuit, however, it is useful to
imagine a charge moving all the
way around the circuit.
radiation, representing another means of escape for the energy. After some time
interval has passed, the resistor reaches a constant temperature, at which time the
input of energy from the battery is balanced by the output of energy by heat and
radiation. Some electrical devices include heat sinks 4 connected to parts of the circuit
to prevent these parts from reaching dangerously high temperatures. These are pieces
of metal with many ﬁns. The high thermal conductivity of the metal provides a rapid
transfer of energy by heat away from the hot component, while the large number of
ﬁns provides a large sur face area in contact with the air, so that energy can transfer by
radiation and into the air by heat at a high rate.
Let us consider now the rate at which the system loses electric potential energy as
the charge Q passes through the resistor:
dU
dt
d
(Q ∆V )
dt
dQ
∆V
dt
I ∆V
where I is the current in the circuit. The system regains this potential energy when the
charge passes through the battery, at the expense of chemical energy in the battery.
The rate at which the system loses potential energy as the charge passes through the
resistor is equal to the rate at which the system gains internal energy in the resistor.
Thus, the power , representing the rate at which energy is delivered to the resistor, is
IV
Power delivered to a device
(27.22)
We derived this result by considering a battery delivering energy to a resistor. However,
Equation 27.22 can be used to calculate the power delivered by a voltage source to any
device carrying a current I and having a potential difference V between its terminals.
Using Equation 27.22 and the fact that V I R for a resistor, we can express the
power delivered to the resistor in the alternative forms
I 2R
Power delivered to a resistor
L
PITFALL PR EVE NTI O N
27.8 Energy Is Not
“Dissipated”
In some books, you may see
Equation 27.23 described as the
power “dissipated in” a resistor,
suggesting that energy disappears.
Instead we say energy is “delivered
to” a resistor. The notion of
dissipation arises because a warm
resistor will expel energy by
radiation and heat, so that energy
delivered by the battery leaves the
circuit. (It does not disappear!)
( V )2
R
(27.23)
When I is expressed in amperes, V in volts, and R in ohms, the SI unit of power is the
watt, as it was in Chapter 7 in our discussion of mechanical power. The process by
which power is lost as internal energy in a conductor of resistance R is often called joule
heating 5; this transformation is also often referred to as an I 2R loss.
When transporting energy by electricity through power lines, such as those
shown in the opening photograph for this chapter, we cannot make the simplifying
assumption that the lines have zero resistance. Real power lines do indeed have
resistance, and power is delivered to the resistance of these wires. Utility companies
seek to minimize the power transformed to internal energy in the lines and maximize the energy delivered to the consumer. Because
I V, the same amount of
power can be transported either at high currents and low potential differences or at
low currents and high potential differences. Utility companies choose to transport
energy at low currents and high potential differences primarily for economic
reasons. Copper wire is ver y expensive, and so it is cheaper to use high-resistance
wire (that is, wire having a small cross-sectional area; see Eq. 27.11). Thus, in the
expression for the power delivered to a resistor,
I 2R , the resistance of the wire
is ﬁxed at a relatively high value for economic considerations. The I 2R loss can be
reduced by keeping the current I as low as possible, which means transferring the
energy at a high voltage. In some instances, power is transported at potential
differences as great as 765 kV. Once the electricity reaches your city, the potential
difference is usually reduced to 4 kV by a device called a transformer. Another
4
This is another misuse of the word heat that is ingrained in our common language.
5
It is commonly called joule heating even though the process of heat does not occur. This is another
example of incorrect usage of the word heat that has become entrenched in our language.
S E C T I O N 2 7. 6 • Electrical Power
847
transformer drops the potential difference to 240 V before the electricity ﬁnally
reaches your home. Of course, each time the potential difference decreases, the
current increases by the same factor, and the power remains the same. We shall
discuss transformers in greater detail in Chapter 33.
Demands on our dwindling energy supplies have made it necessar y for us to be
aware of the energy requirements of our electrical devices. Ever y electrical
appliance carries a label that contains the information you need to calculate the
appliance’s power requirements. In many cases, the power consumption in watts is
stated directly, as it is on a lightbulb. In other cases, the amount of current used by
the device and the potential difference at which it operates are given. This
information and Equation 27.22 are sufﬁcient for calculating the power requirement of any electrical device.
Quick Quiz 27.7 The same potential difference is applied to the two
lightbulbs shown in Figure 27.14. Which one of the following statements is true?
(a) The 30-W bulb carries the greater current and has the higher resistance.
(b) The 30-W bulb carries the greater current, but the 60-W bulb has the higher
resistance. (c) The 30-W bulb has the higher resistance, but the 60-W bulb carries
the greater current. (d) The 60-W bulb carries the greater current and has the
higher resistance.
30 W
e
f
George Semple
60 W
c
d
Figure 27.14 (Quick Quiz 27.7) These lightbulbs operate at their rated power only
when they are connected to a 120-V source.
a
Quick Quiz 27.8
For the two lightbulbs shown in Figure 27.15, rank the
current values at points a through f, from greatest to least.
Example 27.7
Solution Because V
Figure 27.15 (Quick Quiz 27.8)
Two lightbulbs connected across
the same potential difference.
∆V
R
120 V
8.00
(15.0 A)2(8.00
15.0 A
)
1.80
1.80 kW
What If? What if the heater were accidentally connected to a
240-V supply? (This is difﬁcult to do because the shape and
orientation of the metal contacts in 240-V plugs are different
from those in 120-V plugs.) How would this affect the current
carried by the heater and the power rating of the heater?
IR , we have
We can ﬁnd the power rating using the expression
I 2R :
I 2R
b
Power in an Electric Heater
An electric heater is constructed by applying a potential
difference of 120 V to a Nichrome wire that has a total
resistance of 8.00 . Find the current carried by the wire
and the power rating of the heater.
I
∆V
103 W
Answer If we doubled the applied potential difference,
Equation 27.8 tells us that the current would double.
According to Equation 27.23,
( V )2/R, the power
would be four times larger.
848
C HAPTE R 2 7 • Current and Resistance
Linking Electricity and Thermodynamics
Interactive
(A) What is the required resistance of an immersion heater
that will increase the temperature of 1.50 kg of water from
10.0°C to 50.0°C in 10.0 min while operating at 110 V?
the electrical power. The amount of energy transfer by heat
necessary to raise the temperature of the water is given by
Equation 20.4, Q mc T. Thus,
Example 27.8
(B) Estimate the cost of heating the water.
Solution This example allows us to link our new understanding of power in electricity with our experience with
speciﬁc heat in thermodynamics (Chapter 20). An immersion heater is a resistor that is inserted into a container of
water. As energy is delivered to the immersion heater,
raising its temperature, energy leaves the sur face of the
resistor by heat, going into the water. When the immersion
heater reaches a constant temperature, the rate of energy
delivered to the resistance by electrical transmission is equal
to the rate of energy delivered by heat to the water.
(A) To simplify the analysis, we ignore the initial period
during which the temperature of the resistor increases, and
also ignore any variation of resistance with temperature. Thus,
we imagine a constant rate of energy transfer for the entire
10.0 min. Setting the rate of energy delivered to the resistor
equal to the rate of energy entering the water by heat, we have
(∆V )2
R
Q
∆t
(∆V )2
R
mc ∆T
∆t
9:
(∆V )2 ∆t
mc ∆T
R
Substituting the values given in the statement of the
problem, we have
R
(110 V )2(600 s)
(1.50 kg)(4186 J/kg C)(50.0 C
10.0 C)
28.9
(B) Because the energy transferred equals power multiplied
by time interval, the amount of energy transferred is
∆t
(∆V )2
∆t
R
(110 V )2
(10.0 min)
28.9
69.8 W h
1h
60.0 min
0.069 8 kW h
If the energy is purchased at an estimated price of 10.0¢ per
kilowatt-hour, the cost is
Cost
(0.069 8 kWh)($0.100/kWh)
where Q represents an amount of energy transfer by heat
into the water and we have used Equation 27.23 to express
$0.006 98
0.7 ¢
At the Interactive Worked Example link at http://www.pse6.com, you can explore the heating of the water.
Example 27.9
Current in an Electron Beam
In a certain particle accelerator, electrons emerge with
an energy of 40.0 MeV (1 MeV 1.60 10 13 J). The
electrons emerge not in a steady stream but rather in pulses
at the rate of 250 pulses/s. This corresponds to a time
interval between pulses of 4.00 ms (Fig. 27.16). Each pulse
has a duration of 200 ns, and the electrons in the pulse
constitute a current of 250 mA. The current is zero between
pulses.
Solution We use Equation 27.2 in the form dQ I dt and
integrate to ﬁnd the charge per pulse. While the pulse is on,
the current is constant; thus,
(A) How many electrons are delivered by the accelerator
per pulse?
Dividing this quantity of charge per pulse by the electronic
charge gives the number of electrons per pulse:
Q pulse
I
dt
5.00
4.00 ms
I ∆t
10
(250
3
A)(200
10
9 s)
8C
2.00 × 10–7 s
I
10
t (s)
Figure 27.16 (Example 27.9) Current versus time for a pulsed beam of electrons.
Summary
5.00 10 8 C/pulse
1.60 10 19 C/electron
Electrons per pulse
1011 electrons/pulse
3.13
We could also compute this power directly. We assume that
each electron has zero energy before being accelerated.
Thus, by deﬁnition, each electron must go through a
potential difference of 40.0 M V to acquire a ﬁnal energy of
40.0 MeV. Hence, we have
(B) What is the average current per pulse delivered by the
accelerator?
Solution Average current is given by Equation 27.1,
I av
Q / t. Because the time interval between pulses is
4.00 ms, and because we know the charge per pulse from
part (A), we obtain
Q pulse
I av
∆t
5.00
4.00
10
10
8
3
C
s
12.5 A
This represents only 0.005% of the peak current, which is
250 mA.
(C) What is the peak power delivered by the electron beam?
Solution By deﬁnition, power is energy delivered per unit
time interval. Thus, the peak power is equal to the energy
delivered by a pulse divided by the pulse duration:
(1)
(3.13
pulse energy
pulse duration
107 W
(2)
peak
I peak ∆V
(250
10
3
A)(40.0
106 V )
10.0 MW
What If ? What if the requested quantity in part (C) were the
average power rather than the peak power?
Answer Instead of Equation (1), we would use the time
interval between pulses rather than the duration of a pulse:
av
pulse energy
time interval between pulses
(3.13
1011 electrons/pulse)(40.0 MeV/electron)
4.00 10 3 s/pulse
1.60 10
1 MeV
13
J
500 W
1011 electrons/pulse)(40.0 MeV/electron)
2.00 10 7 s/pulse
1.60 10
1 MeV
1.00
peak
849
13 J
Instead of Equation (2), we would use the average current
found in part (B):
av
Iav ∆V
(12.5
10
6
A)(40.0
106 V )
500 W
Notice that these two calculations agree with each other and
that the average power is much lower than the peak power.
10.0 MW
S U M MARY
The electric current I in a conductor is deﬁned as
dQ
dt
I
(27.2)
where dQ is the charge that passes through a cross section of the conductor in a time
interval dt. The SI unit of current is the ampere (A), where 1 A 1 C/s.
The average current in a conductor is related to the motion of the charge carriers
through the relationship
I av
nqvd A
(27.4)
where n is the density of charge carriers, q is the charge on each carrier, vd is the drift
speed, and A is the cross-sectional area of the conductor.
The magnitude of the current density J in a conductor is the current per
unit area:
J
I
A
nqvd
(27.5)
Take a practice test for
this chapter by clicking on
the Practice Test link at
http://www.pse6.com.
850
C HAPTE R 2 7 • Current and Resistance
The current density in an ohmic conductor is proportional to the electric ﬁeld
according to the expression
J
E
(27.7)
The proportionality constant is called the conductivity of the material of which the
conductor is made. The inverse of
is known as resistivity (that is,
1/ ).
Equation 27.7 is known as Ohm’s law, and a material is said to obey this law if the ratio
of its current density J to its applied electric ﬁeld E is a constant that is independent of
the applied ﬁeld.
The resistance R of a conductor is deﬁned as
R
∆V
I
(27.8)
where V is the potential difference across it, and I is the current it carries.
The SI unit of resistance is volts per ampere, which is deﬁned to be 1 ohm ( );
that is, 1
1 V/A. If the resistance is independent of the applied potential
difference, the conductor obeys Ohm’s law.
For a uniform block of material of cross sectional area A and length , the
resistance over the length is
R
(27.11)
A
where is the resistivity of the material.
In a classical model of electrical conduction in metals, the electrons are treated as
molecules of a gas. In the absence of an electric ﬁeld, the average velocity of the
electrons is zero. When an electric ﬁeld is applied, the electrons move (on the average)
with a drift velocity vd that is opposite the electric ﬁeld and given by the expression
vd
qE
me
(27.14)
where is the average time inter val between electron–atom collisions, me is the mass
of the electron, and q is its charge. According to this model, the resistivity of the
metal is
me
nq 2
(27.17)
where n is the number of free electrons per unit volume.
The resistivity of a conductor varies approximately linearly with temperature
according to the expression
0[1
(T
T0)]
(27.19)
where is the temperature coefﬁcient of resistivity and 0 is the resistivity at some
reference temperature T 0.
If a potential difference V is maintained across a circuit element, the power, or
rate at which energy is supplied to the element, is
IV
Because the potential difference across a resistor is given by
the power delivered to a resistor in the form
I 2R
(∆V )2
R
(27.22)
V
IR, we can express
(27.23)
The energy delivered to a resistor by electrical transmission appears in the form of
internal energy in the resistor.
Questions
851
QU ESTIONS
1. In an analogy between electric current and automobile
trafﬁc ﬂow, what would correspond to charge? What would
correspond to current?
9. When the voltage across a certain conductor is doubled,
the current is observed to increase by a factor of three.
What can you conclude about the conductor?
2. Newspaper articles often contain a statement such as
“10 000 volts of electricity surged through the victim’s
body.’’ What is wrong with this statement?
10. In the water analogy of an electric circuit, what corresponds to the power supply, resistor, charge, and potential
difference?
3. What factors affect the resistance of a conductor?
11. Use the atomic theory of matter to explain why the
resistance of a material should increase as its temperature
increases.
4. What is the difference between resistance and resistivity?
5. Two wires A and B of circular cross section are made of the
same metal and have equal lengths, but the resistance of
wire A is three times greater than that of wire B. What is
the ratio of their cross-sectional areas? How do their radii
compare?
6. Do all conductors obey Ohm’s law? Give examples to
justify your answer.
7. We have seen that an electric ﬁeld must exist inside a
conductor that carries a current. How is it possible in view
of the fact that in electrostatics we concluded that the
electric ﬁeld must be zero inside a conductor?
8. A ver y large potential difference is not necessarily
required to produce long sparks in air. With a device
called Jacob’s ladder, a potential difference of about 10 kV
produces an electric arc a few millimeters long between
the bottom ends of two cur ved rods that project upward
from the power supply. (The device is seen in classic
mad-scientist horror movies and in Figure Q27.8.) The
arc rises, climbing the rods and getting longer and
longer. It disappears when it reaches the top; then a new
spark immediately forms at the bottom and the process
repeats. Explain these phenomena. Why does the arc
rise? Why does a new arc appear only after the previous
one is gone?
12. Why might a “good” electrical conductor also be a “good”
thermal conductor?
13. How does the resistance for copper and for silicon change
with temperature? Why are the behaviors of these two
materials different?
14. Explain how a current can persist in a superconductor
without any applied voltage.
15. What single experimental requirement makes superconducting devices expensive to operate? In principle, can
this limitation be overcome?
16. What would happen to the drift velocity of the electrons in
a wire and to the current in the wire if the electrons could
move freely without resistance through the wire?
17. If charges ﬂow very slowly through a metal, why does it not
require several hours for a light to come on when you
throw a switch?
18. In a conductor, changes in the electric ﬁeld that drives the
electrons through the conductor propagate with a speed
close to the speed of light, although the drift velocity of
the electrons is very small. Explain how these statements
can both be true. Does one particular electron move from
one end of the conductor to the other?
19. Two conductors of the same length and radius are
connected across the same potential difference. One
conductor has twice the resistance of the other. To which
conductor is more power delivered?
20. Two lightbulbs both operate from 120 V. One has a power
of 25 W and the other 100 W. Which bulb has higher
resistance? Which bulb carries more current?
21. Car batteries are often rated in ampere-hours. Does this
designate the amount of current, power, energy, or charge
that can be drawn from the battery?
22. If you were to design an electric heater using Nichrome
wire as the heating element, what parameters of the wire
could you vary to meet a speciﬁc power output, such as
1 000 W ?
Figure Q27.8
C HAPTE R 2 7 • Current and Resistance
852
PROBLEMS
1, 2, 3
straightforward, intermediate, challenging
full solution available in the Student Solutions Manual and Study Guide
coached solution with hints available at http://www.pse6.com
computer useful in solving problem
paired numerical and symbolic problems
Section 27.1
Electric Current
radius of cross section A1 is 0.400 cm. (a) What is the
magnitude of the current density across A1? (b) If the
current density across A2 is one-fourth the value across A1,
what is the radius of the conductor at A2?
1. In a particular cathode ray tube, the measured beam
current is 30.0 A. How many electrons strike the tube
screen every 40.0 s?
2. A teapot with a sur face area of 700 cm2 is to be silver
plated. It is attached to the negative electrode of an
electrolytic cell containing silver nitrate (Ag NO3 ). If
the cell is powered by a 12.0-V battery and has a resistance
of 1.80 , how long does it take for a 0.133-mm layer of
silver to build up on the teapot? (The density of silver is
10.5 103 kg/m3.)
3.
Suppose that the current through a conductor
decreases exponentially with time according to the
equation I(t ) I0e t/ where I0 is the initial current (at
t
0), and is a constant having dimensions of time.
Consider a ﬁxed observation point within the conductor.
(a) How much charge passes this point between t 0 and
t
? (b) How much charge passes this point between
t 0 and t 10 ? (c) What If ? How much charge passes
this point between t 0 and t
?
4. In the Bohr model of the hydrogen atom, an electron
in the lowest energy state follows a circular path
5.29 10 11 m from the proton. (a) Show that the speed
of the electron is 2.19 106 m/s. (b) What is the effective
current associated with this orbiting electron?
5. A small sphere that carries a charge q is whirled in a circle
at the end of an insulating string. The angular frequency
of rotation is . What average current does this rotating
charge represent?
6. The quantity of charge q (in coulombs) that has passed
through a sur face of area 2.00 cm2 varies with time
according to the equation q 4t 3 5t 6, where t is in
seconds. (a) What is the instantaneous current through
the sur face at t 1.00 s? (b) What is the value of the
current density?
7. An electric current is given by the expression I (t )
100 sin(120 t), where I is in amperes and t is in seconds.
What is the total charge carried by the current from t 0
to t (1/240) s?
8. Figure P27.8 represents a section of a circular conductor
of nonuniform diameter carrying a current of 5.00 A. The
A2
A1
I
Figure P27.8
9. The electron beam emerging from a certain high-energy
electron accelerator has a circular cross section of radius
1.00 mm. (a) The beam current is 8.00 A. Find the current
density in the beam, assuming that it is uniform throughout.
(b) The speed of the electrons is so close to the speed of
light that their speed can be taken as c
3.00 108 m/s
with negligible error. Find the electron density in the beam.
(c) How long does it take for Avogadro’s number of
electrons to emerge from the accelerator?
10. A Van de Graaff generator produces a beam of 2.00-MeV
deuterons, which are heavy hydrogen nuclei containing a
proton and a neutron. (a) If the beam current is 10.0 A,
how far apart are the deuterons? (b) Is the electric force
of repulsion among them a signiﬁcant factor in beam
stability? Explain.
11. An aluminum wire having a cross-sectional area of
4.00 10 6 m2 carries a current of 5.00 A. Find the drift
speed of the electrons in the wire. The density of
aluminum is 2.70 g/cm3. Assume that one conduction
electron is supplied by each atom.
Section 27.2 Resistance
12. Calculate the current density in a gold wire at 20°C, if an
electric ﬁeld of 0.740 V/m exists in the wire.
13. A lightbulb has a resistance of 240 when operating with
a potential difference of 120 V across it. What is the
current in the lightbulb?
14. A resistor is constructed of a carbon rod that has a
uniform cross-sectional area of 5.00 mm2. When a
potential difference of 15.0 V is applied across the ends of
the rod, the rod carries a current of 4.00 10 3 A. Find
(a) the resistance of the rod and (b) the rod’s length.
15.
A 0.900-V potential difference is maintained across a
1.50-m length of tungsten wire that has a cross-sectional
area of 0.600 mm2. What is the current in the wire?
16. A conductor of uniform radius 1.20 cm carries a current
of 3.00 A produced by an electric ﬁeld of 120 V/m. What
is the resistivity of the material?
17. Suppose that you wish to fabricate a uniform wire out of
1.00 g of copper. If the wire is to have a resistance of
R
0.500 , and if all of the copper is to be used, what
will be (a) the length and (b) the diameter of this wire?
18. Gold is the most ductile of all metals. For example, one
gram of gold can be drawn into a wire 2.40 km long. What
is the resistance of such a wire at 20°C? You can ﬁnd the
necessary reference information in this textbook.
Problems
19. (a) Make an order-of-magnitude estimate of the resistance between the ends of a rubber band. (b) Make an
order-of-magnitude estimate of the resistance between
the ‘heads’ and ‘tails’ sides of a penny. In each case state
what quantities you take as data and the values you
measure or estimate for them. (c) WARNING! Do not try
this at home! What is the order of magnitude of the
current that each would carry if it were connected across
a 120-V power supply?
20. A solid cube of silver (density 10.5 g/cm3) has a mass of
90.0 g. (a) What is the resistance between opposite faces of
the cube? (b) Assume each silver atom contributes one
conduction electron. Find the average drift speed of
electrons when a potential difference of 1.00 10 5 V is
applied to opposite faces. The atomic number of silver is
47, and its molar mass is 107.87 g/mol.
21. A metal wire of resistance R is cut into three equal pieces
that are then connected side by side to form a new wire
the length of which is equal to one-third the original
length. What is the resistance of this new wire?
22. Aluminum and copper wires of equal length are found to
have the same resistance. What is the ratio of their radii?
23. A current density of 6.00 10 13 A/m2 exists in the atmosphere at a location where the electric ﬁeld is 100 V/m.
Calculate the electrical conductivity of the Earth’s atmosphere in this region.
24. The rod in Figure P27.24 is made of two materials. The ﬁgure is not drawn to scale. Each conductor has a square
cross section 3.00 mm on a side. The ﬁrst material has a
resistivity of 4.00 10 3
m and is 25.0 cm long, while
the second material has a resistivity of 6.00 10 3
m
and is 40.0 cm long. What is the resistance between the
ends of the rod?
40.0 cm
25.0 cm
Figure P27.24
Section 27.3 A Model for Electrical Conduction
25.
If the magnitude of the drift velocity of free electrons
in a copper wire is 7.84 10 4 m/s, what is the electric
ﬁeld in the conductor?
26. If the current carried by a conductor is doubled, what
happens to the (a) charge carrier density? (b) current
density? (c) electron drift velocity? (d) average time
interval between collisions?
27. Use data from Example 27.1 to calculate the collision
mean free path of electrons in copper. Assume the average
thermal speed of conduction electrons is 8.60 105 m/s.
Section 27.4 Resistance and Temperature
28. While taking photographs in Death Valley on a day when
the temperature is 58.0°C, Bill Hiker ﬁnds that a certain
voltage applied to a copper wire produces a current of
1.000 A. Bill then travels to Antarctica and applies the
853
same voltage to the same wire. What current does he
register there if the temperature is 88.0°C? Assume that
no change occurs in the wire’s shape and size.
29. A certain lightbulb has a tungsten ﬁlament with a
resistance of 19.0
when cold and 140
when hot.
Assume that the resistivity of tungsten varies linearly with
temperature even over the large temperature range
involved here, and ﬁnd the temperature of the hot
ﬁlament. Assume the initial temperature is 20.0°C.
30. A carbon wire and a Nichrome wire are connected in
series, so that the same current exists in both wires. If the
combination has a resistance of 10.0 k at 0°C, what is the
resistance of each wire at 0°C so that the resistance of
the combination does not change with temperature? The
total or equivalent resistance of resistors in series is
the sum of their individual resistances.
31. An aluminum wire with a diameter of 0.100 mm has a
uniform electric ﬁeld of 0.200 V/m imposed along its
entire length. The temperature of the wire is 50.0°C.
Assume one free electron per atom. (a) Use the information in Table 27.1 and determine the resistivity. (b) What is
the current density in the wire? (c) What is the total
current in the wire? (d) What is the drift speed of the
conduction electrons? (e) What potential difference must
exist between the ends of a 2.00-m length of the wire to
produce the stated electric ﬁeld?
32. Review problem. An aluminum rod has a resistance of
1.234
at 20.0°C. Calculate the resistance of the rod at
120°C by accounting for the changes in both the resistivity
and the dimensions of the rod.
33. What is the fractional change in the resistance of an iron
ﬁlament when its temperature changes from 25.0°C to
50.0°C?
34. The resistance of a platinum wire is to be calibrated for
low-temperature measurements. A platinum wire with
resistance 1.00 at 20.0°C is immersed in liquid nitrogen
at 77 K ( 196°C). If the temperature response of the platinum wire is linear, what is the expected resistance of the
platinum wire at
196°C? ( platinum 3.92 10 3/°C)
35. The temperature of a sample of tungsten is raised while a
sample of copper is maintained at 20.0°C. At what temperature will the resistivity of the tungsten be four times that
of the copper?
Section 27.6 Electrical Power
36. A toaster is rated at 600 W when connected to a 120-V
source. What current does the toaster carry, and what is its
resistance?
37. A Van de Graaff generator (see Figure 25.29) is operating
so that the potential difference between the high-voltage
electrode B and the charging needles at A is 15.0 kV.
Calculate the power required to drive the belt against
electrical forces at an instant when the effective current
delivered to the high-voltage electrode is 500 A.
38. In a hydroelectric installation, a turbine delivers 1 500 hp
to a generator, which in turn transfers 80.0% of the
mechanical energy out by electrical transmission. Under
854
C HAPTE R 2 7 • Current and Resistance
these conditions, what current does the generator deliver
at a terminal potential difference of 2 000 V ?
39.
What is the required resistance of an immersion heater
that increases the temperature of 1.50 kg of water from
10.0°C to 50.0°C in 10.0 min while operating at 110 V ?
40. One rechargeable battery of mass 15.0 g delivers to a CD
player an average current of 18.0 mA at 1.60 V for 2.40 h
before the battery needs to be recharged. The recharger
maintains a potential difference of 2.30 V across the battery
and delivers a charging current of 13.5 mA for 4.20 h.
(a) What is the efﬁciency of the battery as an energy storage
device? (b) How much internal energy is produced in the
battery during one charge–discharge cycle? (b) If the battery
is surrounded by ideal thermal insulation and has an overall
effective speciﬁc heat of 975 J/kg°C, by how much will its
temperature increase during the cycle?
41. Suppose that a voltage surge produces 140 V for a
moment. By what percentage does the power output of a
120-V, 100-W lightbulb increase? Assume that its resistance
does not change.
42. A 500-W heating coil designed to operate from 110 V is
made of Nichrome wire 0.500 mm in diameter. (a) Assuming that the resistivity of the Nichrome remains constant at
its 20.0°C value, ﬁnd the length of wire used. (b) What If ?
Now consider the variation of resistivity with temperature.
What power will the coil of part (a) actually deliver when it
is heated to 1 200°C?
43. A coil of Nichrome wire is 25.0 m long. The wire has a
diameter of 0.400 mm and is at 20.0°C. If it carries a
current of 0.500 A, what are (a) the magnitude of the
electric ﬁeld in the wire, and (b) the power delivered to it?
(c) What If ? If the temperature is increased to 340°C and
the voltage across the wire remains constant, what is the
power delivered?
44. Batteries are rated in terms of ampere-hours (A h). For
example, a battery that can produce a current of 2.00 A
for 3.00 h is rated at 6.00 A h. (a) What is the total
energy, in kilowatt-hours, stored in a 12.0-V battery rated
at 55.0 A h? (b) At $0.060 0 per kilowatt-hour, what is the
value of the electricity produced by this battery?
45. A 10.0-V battery is connected to a 120- resistor. Ignoring
the internal resistance of the battery, calculate the power
delivered to the resistor.
46. Residential building codes typically require the use of
12-gauge copper wire (diameter 0.205 3 cm) for wiring
receptacles. Such circuits carry currents as large as 20 A.
A wire of smaller diameter (with a higher gauge number)
could carry this much current, but the wire could rise to a
high temperature and cause a ﬁre. (a) Calculate the rate
at which internal energy is produced in 1.00 m of 12-gauge
copper wire carrying a current of 20.0 A. (b) What If ?
Repeat the calculation for an aluminum wire. Would a
12-gauge aluminum wire be as safe as a copper wire?
47. An 11.0-W energy-efﬁcient ﬂuorescent lamp is designed to
produce the same illumination as a conventional 40.0-W
incandescent lightbulb. How much money does the user
of the energy-efﬁcient lamp save during 100 hours of use?
Assume a cost of $0.080 0/kWh for energy from the power
company.
48. We estimate that 270 million plug-in electric clocks are in
the United States, approximately one clock for each
person. The clocks convert energy at the average rate
2.50 W. To supply this energy, how many metric tons of
coal are burned per hour in coal-ﬁred electric generating
plants that are, on average, 25.0% efﬁcient? The heat of
combustion for coal is 33.0 MJ/kg.
49. Compute the cost per day of operating a lamp that draws a
current of 1.70 A from a 110-V line. Assume the cost of
energy from the power company is $0.060 0/kWh.
50. Review problem. The heating element of a coffee maker
operates at 120 V and carries a current of 2.00 A.
Assuming that the water absorbs all of the energy delivered to the resistor, calculate how long it takes to raise the
temperature of 0.500 kg of water from room temperature
(23.0°C) to the boiling point.
51. A certain toaster has a heating element made of Nichrome
wire. When the toaster is ﬁrst connected to a 120-V source
(and the wire is at a temperature of 20.0°C), the initial
current is 1.80 A. However, the current begins to decrease
as the heating element warms up. When the toaster
reaches its ﬁnal operating temperature, the current drops
to 1.53 A. (a) Find the power delivered to the toaster when
it is at its operating temperature. (b) What is the ﬁnal
temperature of the heating element?
52. The cost of electricity varies widely through the United
States; $0.120/kWh is one typical value. At this unit
price, calculate the cost of (a) leaving a 40.0-W porch
light on for two weeks while you are on vacation,
(b) making a piece of dark toast in 3.00 min with a
970-W toaster, and (c) dr ying a load of clothes in
40.0 min in a 5 200-W dr yer.
53. Make an order-of-magnitude estimate of the cost of one
person’s routine use of a hair dryer for 1 yr. If you do
not use a blow dryer yourself, observe or interview
someone who does. State the quantities you estimate and
their values.
Additional Problems
54. One lightbulb is marked ‘25 W 120 V,’ and another ‘100 W
120 V’; this means that each bulb has its respective power
delivered to it when plugged into a constant 120-V
potential difference. (a) Find the resistance of each bulb.
(b) How long does it take for 1.00 C to pass through the
dim bulb? Is the charge different in any way upon its exit
from the bulb versus its entry? (c) How long does it take
for 1.00 J to pass through the dim bulb? By what mechanisms does this energy enter and exit the bulb? (d) Find
how much it costs to run the dim bulb continuously for
30.0 days if the electric company sells its product at
$0.070 0 per kW h. What product does the electric company
sell? What is its price for one SI unit of this quantity?
55. A charge Q is placed on a capacitor of capacitance C. The
capacitor is connected into the circuit shown in Figure
P27.55, with an open switch, a resistor, and an initially uncharged capacitor of capacitance 3C. The switch is then
closed and the circuit comes to equilibrium. In terms of Q
and C, ﬁnd (a) the ﬁnal potential difference between the
plates of each capacitor, (b) the charge on each capacitor,
Problems
and (c) the ﬁnal energy stored in each capacitor. (d) Find
the internal energy appearing in the resistor.
C
R
Figure P27.55
56. A high-voltage transmission line with a diameter of 2.00 cm
and a length of 200 km carries a steady current of 1 000 A.
If the conductor is copper wire with a free charge density
of 8.49 1028 electrons/m3, how long does it take one
electron to travel the full length of the line?
57. A more general deﬁnition of the temperature coefﬁcient
of resistivity is
1d
dT
where is the resistivity at temperature T. (a) Assuming
that is constant, show that
(T T0)
where 0 is the resistivity at temperature T0. (b) Using
the series expansion e x 1 x for x
1, show that
the resistivity is given approximately by the expression
(T T0)] for (T T0)
1.
0[1
58. A high-voltage transmission line carries 1 000 A starting at
700 kV for a distance of 100 mi. If the resistance in the
wire is 0.500 /mi, what is the power loss due to resistive
losses?
59.
An experiment is conducted to measure the
electrical resistivity of Nichrome in the form of wires with
different lengths and cross-sectional areas. For one set of
measurements, a student uses 30-gauge wire, which has a
cross-sectional area of 7.30 10 8 m2. The student
measures the potential difference across the wire and the
current in the wire with a voltmeter and an ammeter,
respectively. For each of the measurements given in the
table taken on wires of three different lengths, calculate
the resistance of the wires and the corresponding values of
the resistivity. What is the average value of the resistivity,
and how does this value compare with the value given in
Table 27.1?
L (m)
0.540
1.028
1.543
V (V)
5.22
5.82
5.94
I (A)
R( )
house for a load current of 110 A. For this load current,
ﬁnd (b) the power the customer is receiving and (c) the
electric power lost in the copper wires.
61. A straight cylindrical wire lying along the x axis has a
length of 0.500 m and a diameter of 0.200 mm. It is made
of a material that obeys Ohm’s law with a resistivity of
4.00 10 8
m. Assume that a potential of 4.00 V is
maintained at x 0, and that V 0 at x 0.500 m. Find
(a) the electric ﬁeld E in the wire, (b) the resistance of the
wire, (c) the electric current in the wire, and (d) the
current density J in the wire. Express vectors in vector
notation. (e) Show that E
J.
3C
0e
855
(
m)
0.500
0.276
0.187
60. An electric utility company supplies a customer’s house
from the main power lines (120 V) with two copper wires,
each of which is 50.0 m long and has a resistance of
0.108 per 300 m. (a) Find the voltage at the customer’s
62. A straight cylindrical wire lying along the x axis has a
length L and a diameter d. It is made of a material that
obeys Ohm’s law with a resistivity . Assume that potential
V is maintained at x 0, and that the potential is zero at
x L . In terms of L, d, V, , and physical constants, derive
expressions for (a) the electric ﬁeld in the wire, (b) the
resistance of the wire, (c) the electric current in the wire,
and (d) the current density in the wire. Express vectors in
vector notation. (e) Prove that E
J.
63. The potential difference across the ﬁlament of a lamp is
maintained at a constant level while equilibrium temperature is being reached. It is observed that the steady-state
current in the lamp is only one tenth of the current drawn
by the lamp when it is ﬁrst turned on. If the temperature
coefﬁcient of resistivity for the lamp at 20.0°C is
0.004 50 (°C) 1, and if the resistance increases linearly
with increasing temperature, what is the ﬁnal operating
temperature of the ﬁlament?
64. The current in a resistor decreases by 3.00 A when the
voltage applied across the resistor decreases from 12.0 V to
6.00 V. Find the resistance of the resistor.
65. An electric car is designed to run off a bank of 12.0-V batteries with total energy storage of 2.00 107 J. (a) If the
electric motor draws 8.00 kW, what is the current delivered
to the motor? (b) If the electric motor draws 8.00 kW as
the car moves at a steady speed of 20.0 m/s, how far will
the car travel before it is “out of juice”?
66. Review problem. When a straight wire is heated, its
resistance is given by R R 0[1
(T T0)] according to
Equation 27.21, where is the temperature coefﬁcient of
resistivity. (a) Show that a more precise result, one that
includes the fact that the length and area of the wire
change when heated, is
R
R 0[1
(T
[1
T0)][1
(T
2 (T T0)]
T0)]
where
is the coefﬁcient of linear expansion (see
Chapter 19). (b) Compare these two results for a
2.00-m-long copper wire of radius 0.100 mm, ﬁrst at
20.0°C and then heated to 100.0°C.
67. The temperature coefﬁcients of resistivity in Table 27.1 were
determined at a temperature of 20°C. What would they be at
0°C? Note that the temperature coefﬁcient of resistivity at
20°C satisﬁes
(T T0)], where 0 is the
0[1
resistivity of the material at T0 20°C. The temperature
C HAPTE R 2 7 • Current and Resistance
856
coefﬁcient of resistivity
at 0°C must satisfy the expression
T ], where 0 is the resistivity of the material
0[1
at 0°C.
71. Material with uniform resistivity is formed into a wedge
as shown in Figure P27.71. Show that the resistance
between face A and face B of this wedge is
68. An oceanographer is studying how the ion concentration
in sea water depends on depth. She does this by lowering
into the water a pair of concentric metallic cylinders
(Fig. P27.68) at the end of a cable and taking data to
determine the resistance between these electrodes as a
function of depth. The water between the two cylinders
forms a cylindrical shell of inner radius ra , outer radius rb ,
and length L much larger than rb . The scientist applies a
potential difference V between the inner and outer
sur faces, producing an outward radial current I. Let
represent the resistivity of the water. (a) Find the
resistance of the water between the cylinders in terms of L,
, ra , and rb . (b) Express the resistivity of the water in
terms of the measured quantities L, ra , rb , V, and I.
rb
L
R
w( y 2
Face B
y1
y2
L
w
72. A material of resistivity is formed into the shape of a
truncated cone of altitude h as shown in Figure P27.72.
The bottom end has radius b, and the top end has radius
a. Assume that the current is distributed uniformly over
any circular cross section of the cone, so that the current
density does not depend on radial position. (The current
density does vary with position along the axis of the cone.)
Show that the resistance between the two ends is described
by the expression
69. In a certain stereo system, each speaker has a resistance of
4.00 . The system is rated at 60.0 W in each channel, and
each speaker circuit includes a fuse rated 4.00 A. Is this
system adequately protected against overload? Explain
your reasoning.
70. A close analogy exists between the ﬂow of energy by heat
because of a temperature difference (see Section 20.7)
and the ﬂow of electric charge because of a potential
difference. The energy dQ and the electric charge dq can
both be transported by free electrons in the conducting
material. Consequently, a good electrical conductor is
usually a good thermal conductor as well. Consider a thin
conducting slab of thickness dx, area A, and electrical
conductivity , with a potential difference dV between
opposite faces. Show that the current I dq/dt is given by
the equation on the left below :
A
dV
dx
Thermal Conduction
(Eq. 20.14)
dQ
dt
y2
y1
Figure P27.71
Figure P27.68
dq
dt
ln
Face A
ra
L
Charge Conduction
y 1)
kA
dT
dx
In the analogous thermal conduction equation on the
right, the rate of energy ﬂow dQ /dt (in SI units of joules
per second) is due to a temperature gradient dT/dx, in a
material of thermal conductivity k. State analogous rules
relating the direction of the electric current to the change
in potential, and relating the direction of energy ﬂow to
the change in temperature.
h
ab
R
a
h
b
Figure P27.72
73. The dielectric material between the plates of a parallelplate capacitor always has some nonzero conductivity .
Let A represent the area of each plate and d the distance
between them. Let represent the dielectric constant of
the material. (a) Show that the resistance R and the
capacitance C of the capacitor are related by
RC
0
(b) Find the resistance between the plates of a 14.0-nF
capacitor with a fused quartz dielectric.
74.
The current–voltage characteristic curve for a semiconductor diode as a function of temperature T is given by
the equation
I
I 0(e e ∆V/k BT
1)
Answers to Quick Quizzes
Here the ﬁrst symbol e represents Euler’s number, the base
of natural logarithms. The second e is the charge on the
electron. The k B stands for Boltzmann’s constant, and T is
the absolute temperature. Set up a spreadsheet to
calculate I and R
V/I for V 0.400 V to 0.600 V in
increments of 0.005 V. Assume I0 1.00 nA. Plot R versus
V for T 280 K, 300 K, and 320 K.
75. Review problem. A parallel-plate capacitor consists of
square plates of edge length
that are separated by a
distance d, where d
. A potential difference V is
maintained between the plates. A material of dielectric
constant ﬁlls half of the space between the plates. The
dielectric slab is now withdrawn from the capacitor, as
shown in Figure P27.75. (a) Find the capacitance when
the left edge of the dielectric is at a distance x from the
center of the capacitor. (b) If the dielectric is removed at a
constant speed v, what is the current in the circuit as the
dielectric is being withdrawn?
v
∆V
d
x
Figure P27.75
Answers to Quick Quizzes
27.1 d, b c, a. The current in part (d) is equivalent to two
positive charges moving to the left. Parts (b) and (c)
each represent four positive charges moving in the same
direction because negative charges moving to the left are
equivalent to positive charges moving to the right. The
current in part (a) is equivalent to ﬁve positive charges
moving to the right.
27.2 (b). The currents in the two paths add numerically to
equal the current coming into the junction, without
regard for the directions of the two wires coming out of
857
the junction. This is indicative of scalar addition. Even
though we can assign a direction to a current, it is not a
vector. This suggests a deeper meaning for vectors
besides that of a quantity with magnitude and direction.
27.3 (a). The current in each section of the wire is the same
even though the wire constricts. As the cross-sectional
area A decreases, the drift velocity must increase in order
for the constant current to be maintained, in accordance
with Equation 27.4. As A decreases, Equation 27.11 tells
us that R increases.
27.4 (b). The doubling of the radius causes the area A to be
four times as large, so Equation 27.11 tells us that the
resistance decreases.
27.5 (b). The slope of the tangent to the graph line at a point
is the reciprocal of the resistance at that point. Because
the slope is increasing, the resistance is decreasing.
27.6 (a). When the ﬁlament is at room temperature, its
resistance is low, and hence the current is relatively large.
As the ﬁlament warms up, its resistance increases, and
the current decreases. Older lightbulbs often fail just as
they are turned on because this large initial current
“spike” produces rapid temperature increase and
mechanical stress on the ﬁlament, causing it to break.
27.7 (c). Because the potential difference V is the same
across the two bulbs and because the power delivered to
a conductor is
I V, the 60-W bulb, with its higher
power rating, must carry the greater current. The 30-W
bulb has the higher resistance because it draws less
current at the same potential difference.
27.8 Ia Ib Ic Id Ie If . The current Ia leaves the
positive terminal of the battery and then splits to ﬂow
through the two bulbs; thus, Ia Ic Ie . From Quick
Quiz 27.7, we know that the current in the 60-W bulb is
greater than that in the 30-W bulb. Because charge does
not build up in the bulbs, we know that the same amount
of charge ﬂowing into a bulb from the left must ﬂow out
on the right; consequently, Ic Id and Ie If . The two
currents leaving the bulbs recombine to form the current
back into the battery, If Id Ib .
...

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